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T-shirt launcher and Projectile motion

  1. Feb 13, 2017 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Hello,

    One more question, this time it's about projectile motion. Could someone please check my answer to this problem. All my work and applicable equations are shown in the pictures below. If my answer is wrong, please provide me with guidelines for solving the problem. I'm not expecting anyone to just give me the answer :)

    Thanks

    Problem 9

    You're attending a professional basketball game. The home team calls timeout, and, as the music gets cranked up, the t-shirt crew comes out onto the floor to fire team t-shirts into the crowd. Like everyone else, you stand up and wave your arms wildly, hoping that one of the t-shirt guys will fire a shirt your way. The t-shirts are fired from a launcher that uses compressed air. One of the launchers is pointed your way, and is just about to launch a t-shirt at a 45° angle from the horizontal. You are a horizontal distance of 16.0 m from the launcher and a vertical distance of 6.00 m above it. Neglect air resistance, and assume that the acceleration due to gravity is 9.80 m/s2.

    (a) What is the launch speed needed so that the t-shirt reaches you, rather than one of the hordes of people sitting near you?

    (b) What is the maximum height reached by the t-shirt (measured from the level of the launcher)?

    (c) How long does the t-shirt spend in the air before reaching you?

    (d)What is the speed of the t-shirt when it reaches you?

    IMG_3091.JPG IMG_3092.JPG IMG_3093.JPG IMG_3094.JPG IMG_3095.JPG
     
  2. jcsd
  3. Feb 13, 2017 #2

    berkeman

    User Avatar

    Staff: Mentor

    It would be better if you could type your work into the forum window. At ;lease your images are fairly readable...
     
  4. Feb 13, 2017 #3
    Sorry, I'll remember to type them up next time. Its just that I'm in a little bit of a rush and don't have time to type out all of my work. I tried to present them as clearly as possible in the pictures. Do you think you could still help me based on this? If you would still prefer me to type out the equations I will try my best to do that.
     
  5. Feb 13, 2017 #4
    I haven't worked my way through your entire solution, but in part a) you assumed that the y component of the final velocity would be 0. Why is that?

    Edit: Also, 51 m/s is about 115 mph. That just seems kind of high.
     
    Last edited: Feb 13, 2017
  6. Feb 13, 2017 #5

    Student100

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    Education Advisor
    Gold Member

    He does have pretty handwriting, at least compared to mine. I hope he reads my previous post to his homework question and starts trying to work things out algebraically before plugging in numbers though.

    OP If you have time https://www.physicsforums.com/help/latexhelp/ for when you start typing out things... most of what you will need is \frac{a}{b} which gives you ##\frac{a}{b}## and x_0 which would be ##x_0## or x^2 which gives ##x^2##. Then you just surround those things with #*# #*# (in text) or $*$ $*$ (new line) (both without the *) . Hit preview to make sure you're typing things correctly.

    It will also be useful for more than just this forum! Most everyone uses latex the further you go in the sciences. PM if you need additional help with typing stuff out. (Not for homework, keep that here)

    I'm eating dinner, but will work this out later if someone doesn't beat me to it. :nb)
     
  7. Feb 13, 2017 #6
    I thought that the final velocity would be zero upon reaching the person in the crowd. Dosen't the launcher give the t-shirt an intital velocity?
     
  8. Feb 13, 2017 #7
    ok, thanks!
     
  9. Feb 13, 2017 #8

    gneill

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    Staff: Mentor

    Suppose you throw a ball at a wall. Does the ball have zero velocity when it hits the wall?
     
  10. Feb 14, 2017 #9
    The launcher gives the shirt an initial velocity, yes. But there is nothing in the problem that indicates the vertical component of velocity is 0 m/s when it reaches the person at the 6 m height. That was not a valid assumption. You should really expect that the shirt reaches a height of 6 meters at two different times - once on the way up and once on the way down. It will have a vertical velocity of 0 at the peak though.
     
  11. Feb 14, 2017 #10
    no
     
  12. Feb 14, 2017 #11
    You know, I was looking at your solution again and I just realized a couple of other problems.
    On the 2nd page, in trying to solve part a), you wrote the following consecutive equations:
    vi2 = 235.2
    viy = 15.34 m/s
    So you were solving for the overall velocity all along and then at the end changed it to the vertical component of the velocity.

    After that you solved for the time. That quadratic equation has 2 solutions. You found one of the solutions = 0.458 s. As I mentioned previously, the shirt will be at a height of 6 meters at two times - once on the way up and once on the way down. The shorter time will be when it hits that height on the way up, and the longer time will be when it reaches that height on the way down.

    One more thought on part a. Here are your knowns: Horizontal distance = 16 m, vertical distance = 6 m, angle = 45°, gravity = -9.8 m/s.
    And like you did, write equations for x and y components. But, those equations will have to be solved simultaneously because there is not enough information for either component (x or y) to solve it independently. So for the x component, you basically have x = vt (because as you showed, all the other terms are 0). The only thing you know is x. So you will solve for either v in terms of t, or t in terms of v, and then plug that into whatever equation you use for the y component.
     
  13. Feb 15, 2017 #12
    Thanks TomHart. I went back and looked over my work and also realized that I made a bunch of mistakes. I redid the whole problem; would you (or any other physics expert) mind checking out my solutions please? I'm feeling pretty confident about A through C, but not so much about D though.


    IMG_3096.JPG IMG_3097.JPG IMG_3099.JPG
     
  14. Feb 15, 2017 #13
    a), b) and c) look right. d) has a problem. You know the initial velocity and you know the launch angle. Therefore you should be able to find the x component of the initial velocity easily enough. And what is the acceleration in the x direction? So it should be easy to find the x component of the final velocity. It's the same as the initial. So you no longer need to solve 2 simultaneous equations. You should be able to calculate the x component and the y component of the velocity independently now.

    Your problem with the y component calculation is this: In your equation for the x component time, you calculated how long it took to get from x = 8 to x = 16. That is half the x overall distance. But the maximum height is not reached at the midpoint of the x distance. So that is causing an error in your calculations.

    There are 3 knowns in the y direction. You know the initial velocity, you know the time it takes to reach the final destination, and you know the acceleration. Knowing those three, you are able to calculate the y component final velocity.

    Once you know the final x component velocity and final y component velocity, it's a simple matter to find the overall velocity. <-- Edit: I mean speed, not velocity.
     
  15. Feb 15, 2017 #14
    I feel that it would not be forthright for me to let this slip by without commenting. I have been on this forum for less than a year. During that time I have seen a number of very brilliant men and women on here who provide help on homework problems. Those folks certainly qualify as "expert". I do not. I consider myself the resident physics hack. But I am trying to learn. :)
     
  16. Feb 16, 2017 #15
    Thanks for your honesty :) but whether or not you consider yourself an expert, I still trust and value the helpful suggestions you've given me thus far.

    Ok, I re-did part D. Would you say it's correct now?

    IMG_3100.JPG
     
  17. Feb 16, 2017 #16
    Yes, looks right. Good job!
     
  18. Feb 16, 2017 #17
    Thanks!
     
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